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Prove log(n)/n^sigma is a null sequence.

  1. Jul 7, 2015 #1
    Mod note: Moved from the homework section.
    1. The problem statement, all variables and given/known data

    If σ > 0 and base b > 1 then prove that ##log n/n^σ## is a null sequence. This is not really a homework since i am self studying Konrad Knopp book about infinite series and i wanted to see different ideas and perspectives on the proof of the problem.Thanks.
    Last edited by a moderator: Jul 7, 2015
  2. jcsd
  3. Jul 7, 2015 #2


    Staff: Mentor

    Can you provide your proof of the problem?

    Also what is the b value? Is it the base of the log function?
  4. Jul 7, 2015 #3
    Yes the b is the base of the log function.
  5. Jul 7, 2015 #4
    Mod note: Edited text below to make it more readable.
    If b > 1 then we have ##b^σ > 1##. Therefore ##(n/(b^σ)^n)## is a null sequence.
    Given ε > 0 we have consequently from a point onwards - say for every n > m, ##(n/(b^σ)^n) < ε' = ε/b^σ##. But in any case, ##log n/n^σ < b^σ(g + 1/(b^σ)^g b^1 ))## . If g denotes the characteristic of log n . If therefore, ## n > b^m## Then, ## |(log\ n/n^σ)| < ε ∀\ n > b^m## Q.E.D

    This is the proof of the book and it is a theorem not in the exercises as the author is building up the knowledge of the reader, but as i read this again i have one question now how did he come up with what n has to be bigger than ?
    Last edited by a moderator: Jul 7, 2015
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